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The preceding constructions lead to the Iwasawa theorem on local cyclotomic units. Its significance is that an explicitly calculated local unit quotient is governed by a p-adic L-function; class field theory then suggests that the same analytic object should govern global arithmetic. The latter prediction is the Iwasawa main conjecture, rather than a consequence of the local calculation alone.

Fix compatible roots of unity ${\zeta }_{p^n}$ and put $K_n={\mathbb{Q}}_p({\zeta }_{p^n})$, ${\pi }_n={\zeta }_{p^n}-1$. Let $V_{\infty }$ be the norm limit of the local principal units. Take its even part under complex conjugation to obtain $U_{\infty }^1$, the local principal-unit limit in the real tower. Let $C_{\infty }^1$ be the closed subgroup generated by norm-compatible real cyclotomic units, with their finite-order residue factors removed. WriteHere the last map is augmentation. Normalize the p-adic zeta pseudomeasure ${\zeta }_p$ byThe expression $x^k$ is well defined on $G$ for even $k$. A pseudomeasure lies in the total fraction ring of the Iwasawa algebra; the pole on its trivial finite-character component is removed by the augmentation ideal, so $I(G){\zeta }_p\subset {\Lambda }_G$. In this normalization the local theorem isThe following steps outline its proof and explain where the earlier questions enter.

First encode local units by Coleman power series. Evaluation at ${\pi }_n$ is surjective from $R$ to ${\mathcal{O}}_{K_n}$, and the product defining the Coleman norm operator givesThus norm-fixed series give norm-compatible sequences of local units. Conversely, lift $u_m$ from such a sequence to $f^{(m)}\in R^{\times }$ and take $g_m=N^{m-r_m}{f}^{(m)}$, where $r_m=\lfloor m/2\rfloor$. This has $g_m({\pi }_{r_m})=u_{r_m}$. The estimates in Question 2 give $N{g}_m/g_m\in 1+p^{m-r_m+1}R$. Consequently for fixed $n\le r_m$, the series $N^{r_m-n}{g}_m$ differs from $g_m$ modulo $p^{m-r_m+1}$ and its value at ${\pi }_n$ is $u_n$. The compactness of $R$ in the coarser of the topologies on integral formal power series supplies a convergent subsequence. Continuity of the Coleman norm operator and of evaluation shows that its limit is norm-fixed and takes every value $u_n$. It is a unit because the original principal units have residue $1$. Uniqueness follows from the Weierstrass preparation theorem: a nonzero series in $R$ has only finitely many zeros in the open unit disk, whereas the distinct ${\pi }_n$ give infinitely many interpolation points. This constructs the Coleman power series $f_u$.

Next turn its multiplicative information into an integral p-adic measure. For a norm-fixed unit $f$, use the Frobenius-corrected logarithmThe ratio belongs to $1+pR$ by reduction modulo $p$, and the logarithm divided by $p$ is integral because $p$ is odd. Equivalently $\mathcal{L}(f)=(1-\varphi /p)\log f$, interpreting the constant coefficient with the p-adic logarithm on units. The averaging identity for the logarithm is understood on the open $p$-adic unit disk, where the translated logarithmic series converge; the corrected result itself lies in $R$. The logarithmic product formula and Question 1 giveUnder the Amice transform, the root-of-unity average projects a p-adic measure onto its restriction to $p{\mathbb{Z}}_p$: the averaged factor $(1+T{)}^x$ survives exactly for $x\in p{\mathbb{Z}}_p$. Consequently $S\mathcal{L}(f)=0$ says that $\lambda (\mathcal{L}(f))$ is supported on ${\mathbb{Z}}_p^{\times }$. This gives a Galois group-equivariant homomorphism from local principal units to the Iwasawa algebra of ${\mathbb{Z}}_p^{\times }$.

The integral lifting step in this proof is the Coleman unit-measure exact sequenceHere $\mathrm{Col}(u)=\lambda (\mathcal{L}(f_u))$. The last map is surjective, since $m_1({\delta }_1)=1$, and it has the cyclotomic action on its target. Its composition with $\mathrm{Col}$ is zero: $D\varphi =p\varphi D$ implies $D\mathcal{L}(f)(0)=0$. The kernel on the unit side consists of norm-compatible $p$-power roots of unity. To see the shape of the kernel, write $z=\log (1+T)$; the equation $\mathcal{L}(f)=0$ becomes $h(pz)=ph(z)$ for $h(z)=\log f(e^z-1)$, so only its linear term can survive. Integral solutions are $f=(1+T{)}^c$, $c\in {\mathbb{Z}}_p$, after excluding finite residue factors by the principal unit condition. For the image, use the following explicit integral lifting of the corrected cyclotomic logarithm.

Take a p-adic measure $\mu$ on the units with first moment zero, and write $g$ for its Amice transform. Then $Sg=0$ and $g^{\prime }(0)=0$. Put $H=\exp (pg)\in 1+pR$ and solve $f^p=H\varphi (f)$ with $f(0)\in 1+p{\mathbb{Z}}_p$. The constant equation is solved by $f_0=\exp (p{g}_0/(p-1))$. In degree one the terms involving $f_1$ cancel, and the remaining condition is $H_1=0$, exactly $g_1=0$; choose any $f_1\in {\mathbb{Z}}_p$. For $n\ge 2$, after fixing the lower coefficients, the coefficient multiplying the unknown $f_n$ isThe other terms are divisible by $p$: for the already integral truncated series, $f^p\equiv \varphi (f)(\mathrm{mod}p)$, while $H\equiv 1(\mathrm{mod}p)$. Division therefore gives an integral $f_n$, since $f_0$ and $1-p^{n-1}$ are units. Recursion constructs $f\in R^{\times }$ with $\mathcal{L}(f)=g$. Now $pSg=\log (Nf/f)=0$, and $Nf/f\in 1+pR$ by Question 2. The p-adic logarithm is injective there, so $Nf=f$. This supplies the desired Coleman power series of a norm-compatible sequence of local units. The freely chosen linear coefficient accounts precisely for the kernel $(1+T{)}^c$. It proves the integral surjectivity step, including the integrality that rational inversion of $1-\varphi /p$ would leave unproved.

Complex conjugation acts by $-1$ on both copies of ${\mathbb{Z}}_p(1)$. As $p$ is odd, taking the even part is exact and removes both end terms. It follows that $\mathrm{Col}:U_{\infty }^1\simeq {\Lambda }_G$. The integral exact sequence is essential here: the support calculation alone would not prove this isomorphism.

It remains to calculate the image of the cyclotomic units. For $a,b\in {\mathbb{Z}}_p^{\times }$, consider the symmetric cyclotomic unit seriesThe quotient has an invertible constant term $a/b$, because both numerator and denominator have a simple zero at $T=0$. The factor $X^{(b-a)/2}$ makes the expression unchanged under $X\mapsto X^{-1}$. The product over the roots of unity gives $N{f}_{a,b}=f_{a,b}$; one may verify this first for integer $a,b$ prime to $p$ and then extend by continuity. Dividing by the norm-compatible Teichmuller representative makes the series correspond to principal units and does not change its logarithm. The generating function for the Bernoulli numbers, with $z=\log X$, givesThe symmetric factor cancels the linear term. Applying $1-\varphi /p$, and then the moment formula of Question 3, yieldsfor even $k\ge 2$. These moments determine an even p-adic measure on the units: polynomials are dense by the Mahler theorem, even projection preserves density, and multiplication by $x^2$ is invertible on the units, so the positive even-degree polynomials are dense there as well. ThusThe closed ideal generated by all differences $[b]-[a]$ is $I(G)$. The closed subgroup generated by the corresponding cyclotomic units, including its Galois group translates, therefore has image exactly $I(G){\zeta }_p$. Taking the quotient of the isomorphism $U_{\infty }^1\simeq {\Lambda }_G$ proves the boxed local theorem.

Finally introduce global arithmetic. Let ${\overline{E}}_{\infty }^1$ be the closure of norm-compatible global units in the real tower, let $X_{\infty }^{+}$ be its p-ramified Iwasawa module, and let $Y_{\infty }^{+}$ be its unramified Iwasawa module. With compatible unit and splitting conventions, class field theory gives the class-field unit sequence; quotienting its unit terms by $C_{\infty }^1$ givesIn the cyclotomic tower the prime above $p$ is principal, so the unramified class-field convention also makes it split. On the torsion character components, multiplicativity of characteristic ideals impliesThe local theorem has already calculated the first factor analytically. Equality of the global unit and class-group factors would therefore identify the characteristic ideal of $X_{\infty }^{+}$ with the same ideal of the p-adic L-function. That is the content predicted by the Iwasawa main conjecture. More explicitly, for a nontrivial even finite character $\chi$, the character component of $I(G)$ is the whole component and the prediction isHere ${\Lambda }_{\chi }$ is the power-series algebra of the pro-$p$ part with coefficients in the character-value ring. The trivial component requires the augmentation correction for the zeta pole. Equivalently the characteristic ideals of ${\overline{E}}_{\infty }^1/C_{\infty }^1$ and $Y_{\infty }^{+}$ agree componentwise. A formulation using odd class-group characters must include the appropriate cyclotomic twist and inversion, as in Kummer reflection in Iwasawa theory. The motivation is an analytic formula for local units together with a global class-field exact sequence; the global equality is the additional arithmetic assertion.

The related class-number growth theorem illustrates why passage to an Iwasawa module is powerful. For the cyclotomic ${\mathbb{Z}}_p$-extension, let $p^{e_n}$ be the $p$-part of the class number at level $n$. The unramified Iwasawa torsion theorem, the Iwasawa module structure theorem and the class-group control relations give, for sufficiently large $n$,Here $\mu ,\lambda$ are the Iwasawa invariants. To see the two growth terms, use a topological generator $\gamma$ and, after a fixed sufficiently high level $n_0$, the norm polynomial ${\nu }_{n,n_0}=({\gamma }^{p^n}-1)/({\gamma }^{p^{n_0}}-1)$. An elementary summand $\Lambda /(p^a)$ contributes $a(p^n-p^{n_0})$ to the length of its quotient by this polynomial. A summand associated with a distinguished polynomial of degree $d$ contributes $dn$ plus a constant: at each root $\alpha$ the valuation of the norm polynomial is $n$ plus a fixed constant for large $n$. If $1+\alpha$ is a root of unity already killed at level $n_0$, its value is instead exactly $p^{n-n_0}$ and gives the same linear contribution. The finite errors in the structure and control maps stabilize and contribute to $\nu$. Thus module structure explains growth, while the Iwasawa main conjecture goes further by identifying the characteristic series of an Iwasawa module themselves with analytic ones.

For a profinite abelian group $G$, its Iwasawa algebra is the completed group algebrawhere $U$ runs over open subgroups. The transition map sums the coefficients over the fibers of $G/V\to G/U$ when $V\subset U$. Consequently an element gives compatible values $\mu (gU)\in {\mathbb{Z}}_p$ on cosets, equivalently a finitely additive ${\mathbb{Z}}_p$-valued p-adic measure on the clopen sets of $G$. Conversely these values specify every finite-quotient coefficient. Integration of a continuous function with values in ${\mathbb{Z}}_p$ is obtained by uniform approximation by locally constant functions; the integrals converge because all measure values have $p$-adic absolute value at most $1$. Multiplication in the Iwasawa algebra becomes convolution of p-adic measures. This is its measure realization of an Iwasawa algebra.

The Mahler theorem states that each continuous function $g:{\mathbb{Z}}_p\to {\mathbb{Z}}_p$ has the unique expansion converging in the sense of uniform convergenceIts Mahler coefficients areand $\Vert g{\Vert }_{\infty }={\sup }_n|b_n{|}_p$. The same assertion holds for continuous functions with values in ${\mathbb{Q}}_p$ with coefficients in ${\mathbb{Q}}_p$.

Given $f(T)={\sum }_{n\ge 0}{a}_n{T}^n\in R$, define the bounded linear functionalThe sum converges because $a_n\in {\mathbb{Z}}_p$ and $b_n\to 0$. Its values on indicator functions of clopen sets give a p-adic measure, denoted $\lambda (f)$, withConversely, a p-adic measure determines these coefficients $a_n\in {\mathbb{Z}}_p$, and continuity allows termwise integration of every Mahler expansion. Thus the inverse iswhich is the Amice transform. These two constructions are inverse, proving the canonical bijection $\lambda :R\simeq \Lambda ({\mathbb{Z}}_p)$. In fact it is an isomorphism of rings: the identity $\left(\genfrac{}{}{0ex}{}{x+y}{n}\right)={\sum }_{i+j=n}\left(\genfrac{}{}{0ex}{}{x}{i}\right)\left(\genfrac{}{}{0ex}{}{y}{j}\right)$ identifies convolution of p-adic measures with multiplication of formal power series. For example, $\lambda (1)={\delta }_0$ and $\lambda (1+T)={\delta }_1$, where ${\delta }_a$ is the Dirac measure at $a$.

For the moment identity, let $\left\{\genfrac{}{}{0ex}{}{k}{n}\right\}$ denote a Stirling number of the second kind. The finite polynomial identitygives $\int x^{k}d\lambda (f)={\sum }_{n=0}^k{a}_{n}n!\left\{\genfrac{}{}{0ex}{}{k}{n}\right\}$. On the other hand, $D=(1+T)d/dT$ satisfies $D^k(1+T{)}^x=x^k(1+T{)}^x$ for every nonnegative integer $x$. Expanding $(1+T{)}^x$ and evaluating at $T=0$ yieldsComparison in the basis of binomial polynomials gives $(D^k{T}^n)(0)=n!\left\{\genfrac{}{}{0ex}{}{k}{n}\right\}$. Terms with $n>k$ have zero constant term after at most $k$ differentiations, so no infinite-series interchange is needed. ThereforeThe case $k=0$ says that the total mass is $f(0)$.

Let $M_f$ be the matrix of multiplication by $f$ in the basis from Question 1. Define the Coleman norm operator by the norm for a finite free ring extension:The entries lie in $\varphi (R)$, and $\varphi :R\to \varphi (R)$ is an isomorphism, so this determines $Nf$ uniquely. If $f$ is a unit, $M_f$ is invertible and its determinant is a unit. Also $M_{fg}=M_f{M}_g$, giving $N(fg)=NfNg$. Splitting the conjugates, or diagonalizing multiplication after adjoining the roots of unity and inverting $p$, gives the equivalent formulaIn particular, this definition has the required target $R^{\times }$, rather than merely a ring with extra roots of unity in its coefficients.

For the requested ring congruence, observe that the reduction of $\varphi$ is the injective map $\overline{g}(T)\mapsto \overline{g}(T^p)$. If $\varphi (g)\in p^{k}R$, its reduction first gives $g=p{g}_1$. Then $\varphi (g_1)\in p^{k-1}R$; repeat to obtain $g\in p^{k}R$. Applying this to $g=f-1$ provesThis congruence reflection for Frobenius substitution is valid for all $f\in R$, without assuming $f$ is a unit.

A further property needed for the fixed points is Coleman norm contraction. If $g=1+p^{k}h$ with $k\ge 1$, expand the determinant:Here $e_j(M_h)\in \varphi (R)$ is the coefficient of degree $j$ in $\det (I+t{M}_h)$. The trace formula from Question 1 makes the linear term divisible by $p^{k+1}$; every later term is divisible by $p^{2k}$ and $2k\ge k+1$. Reflecting the resulting ring congruence through $\varphi$ givesIf $Nf=f$ and $f\in 1+pR$, iteration gives $f\in 1+p^{k}R$ for every $k$. Since ${\bigcap }_k{p}^{k}R=0$, $f=1$.

To identify all fixed points, we also need $Nf\equiv f(\mathrm{mod}pR)$ for every unit $f$. In characteristic $p$, the extension ${\mathbb{F}}_p[[T]]/{\mathbb{F}}_p[[T^p]]$ has degree $p$ and its norm for a finite free ring extension is $\overline{f}\mapsto {\overline{f}}^p$. Indeed, after passing to fraction fields it is a purely inseparable field extension of degree $p$, so multiplication has only the eigenvalue $\overline{f}$, with multiplicity $p$, after splitting. Its determinant is ${\overline{f}}^p$. Thereforeand congruence reflection for Frobenius substitution gives $Nf\equiv f(\mathrm{mod}pR)$.

Now choose any lift $f_0\in R^{\times }$ of a given $a\in A^{\times }$ and put $f_n=N^n{f}_0$. Multiplicativity givesThus $(f_n)$ is a Cauchy sequence for the finer of the topologies on integral formal power series and converges to a unit $w(a)$ reducing to $a$. The Coleman norm operator is continuous: multiplication matrices and their determinants are continuous, and the finite-free coordinate decomposition and its inverse preserve the $p$-adic filtration defining the finer of the topologies on integral formal power series. Hence $Nw(a)=w(a)$. If two fixed units reduce to $a$, their quotient is fixed and belongs to $1+pR$, so it is $1$ by the preceding argument. This proves uniqueness, independence of the initial lift, and multiplicativity of the Coleman norm-fixed lift. In particular,Every unit $f\in R$ consequently has a unique factorization $f=w(\overline{f})v$ with $v\in 1+pR$, giving the canonical Coleman decomposition of power-series units:The final assertion as printed is false. The original PDF identifies a ring with a product of unit groups. With multiplication, $A$ has a zero and is not a group. With addition, it has exponent $p$, whereas $W\times A^{\times }$ has a nonidentity element of order $2$: $-1\in W$ because $p$ is odd and $N(-1)=(-1{)}^p=-1$. Thus neither interpretation makes the printed assertion true. The two boxed isomorphisms above are the valid natural conclusions; they do not require guessing which symbols the author intended to replace.

Put $X=1+T$ and $Y=X^p-1$, so that $\varphi (f)=f(Y)$. The essential integrality statement for the Frobenius substitution on cyclotomic power series is the decomposition into a finite free module:To prove it, reduce modulo $p$. Since $Y\equiv T^p$, grouping the coefficients of a formal power series according to their exponents modulo $p$ gives the unique expression ${\sum }_{j=0}^{p-1}{T}^j{h}_j(T^p)$. The change from $1,T,\dots ,T^{p-1}$ to $1,X,\dots ,X^{p-1}$ is a triangular matrix with diagonal entries $1$, so this is also a basis over ${\mathbb{F}}_p[[T^p]]$. Lift these $p$ coefficient series to $R$, subtract their contribution from $f$, divide the remaining error by $p$, and repeat. The sums of the successive lifts converge in the finer of the topologies on integral formal power series, because $R$ is complete in that topology. This proves existence of the decomposition over $R$. A relation among its summands reduces to the zero relation modulo $p$, forcing each coefficient series to be divisible by $p$. Repetition forces divisibility by every power of $p$, hence every coefficient series is zero. This proves uniqueness and, in particular, injectivity of $\varphi$.

For $f={\sum }_{i=0}^{p-1}{X}^i\varphi (f_i)$, defineThis is a ${\mathbb{Z}}_p$-linear map. To calculate the average over roots of unity, work over $\mathcal{O}={\mathbb{Z}}_p[{\zeta }_p]$. The substitutions $T\mapsto \zeta X-1$ are legitimate: their constant terms are topologically nilpotent, so the defining coefficient sums converge in $\mathcal{O}$. Since $(\zeta X{)}^p=X^p$, each $\varphi (f_i)$ is unchanged by the substitution. The sums of the powers of the roots of unity are $p$ for $i=0$ and zero for $1\le i<p$. ConsequentlyThus the average is integral and belongs to $\varphi (R)$, and the constructed Coleman trace operator has the required property. Injectivity of $\varphi$ forces uniqueness of any map satisfying that property. Finally, the decomposition of $\varphi (g)$ has $f_0=g$ and all other coefficients zero, giving $S\circ \varphi ={\mathrm{id}}_R$.

We will also use the trace of multiplication by $h$ on this finite free module. After extending scalars to split the conjugates, that trace is the sum just computed. Hence

Choose $m\ge \ell (\lambda )$, pad the partition of an integer with zeros, and define its beta numbers byThey form a strictly decreasing set of nonnegative integers. Conversely, from its decreasing enumeration recover ${\lambda }_i={\beta }_i-m+i$. With $m=\ell (\lambda )$ these are exactly the first-column hook lengths, since $h_{i1}={\lambda }_i+{\lambda }_1^{\prime }-i={\lambda }_i+m-i$. Padding changes the numbers: increasing $m$ by one replaces $B$ by $\{0\}\cup (B+1)$, and leaves the represented partition of an integer unchanged.

Arrange positions $0,1,2,\dots$ on $p$ runners, putting position $r+pk$ on runner $r$ at height $k$. A bead occupies each position in the beta set. This is the abacus of a partition. The hook criterion in a beta set says that a bead $b$ and a gap $a<b$ correspond to a hook of length $b-a$; replacing $b$ by $a$ removes the corresponding rim hook. In particular removing a rim $p$-hook moves a bead from $b$ to the vacant position $b-p$ immediately above it on the same runner. Its leg length counts the beads in the intervening positions of the full linear numbering.

For an explicit example, take $\lambda =(4,2,1)$, $p=3$, and $m=3$. Its beta set is $\{6,3,1\}$. Runner zero has beads at $3,6$, runner one has a bead at $1$, and runner two is empty. Packing upwards gives $\{3,1,0\}$, which represents $(1)$; thus the three-core is $(1)$ and the weight is two.

Every bead slide reduces the sum of occupied positions, so repeated slides terminate. The number $n_r$ of beads on each runner is unchanged. At termination they must occupy exactlyon runner $r$, since any gap above a bead would allow a further slide. Hence the final beta set is uniquely determined by the initial runner counts, independently of the order of hook removals. Increasing the number of beads by $p$ replaces $B$ by $\{0,\dots ,p-1\}\cup (B+p)$, adds one bead to each runner, and gives exactly the same represented core after packing. General changes in padding follow by applying $B\mapsto \{0\}\cup (B+1)$; packing commutes with this operation up to the cyclic relabelling of runners. Therefore the result is independent of both removal order and padding:The weight of a partition is $w_p(\lambda )=(n-|C_p(\lambda )|)/p$.

A block of a group algebra is an indecomposable two-sided summand $F{S}_{n}e$, where $e$ is a primitive central idempotent. The identity decomposes into pairwise orthogonal such idempotents; a module belongs to a block when its idempotent acts as identity and the other block idempotents act as zero. For ordinary characters, use a splitting p-modular system $(K,\mathcal{O},F)$, or extend the residue field first: the block idempotents lift to $\mathcal{O}{S}_n$, and determine which ordinary irreducible $K{S}_n$-modules lie in each $p$-block. The symmetric group is split over its prime field, so extending $F$ introduces no change to the resulting partition of an integer labels.

The Nakayama block theorem, traditionally called the Nakayama Conjecture for symmetric groups, isThe simple modules in that block are exactly the $D^{\nu }$ with regular labels having that core. At fixed $n$, the common core also fixes the weight. Here is an essay proof route through central characters, specifying the main algebraic and combinatorial steps and why each is needed.

First introduce the commuting Young–Jucys–Murphy elementsTheir commutators vanish because contributions supported on each triple of letters cancel. Symmetric polynomials in the $J_i$ are central: for adjacent transpositions $s_i$, the relations $s_i{J}_i=J_{i+1}{s}_i-1$, $s_i{J}_{i+1}=J_i{s}_i+1$ show that $s_i$ commutes with symmetric expressions in $J_i,J_{i+1}$, and it commutes with the other $J_j$. Since these transpositions generate $S_n$, the symmetric expressions commute with the whole group algebra.

The substantial generation ingredient is the integral Jucys–Murphy description of the center of a symmetric-group algebra: every conjugacy-class sum is an integral symmetric polynomial in the $J_i$. Its proof expands symmetric monomials in these sums of transpositions and organizes the resulting class sums by the number of moved letters and cycle filtration; triangular elimination supplies integral spanning, rather than merely rational spanning. A useful first identity in this construction isIndeed, when adjoining the letter $n$, either choose $t$ and leave it as a singleton cycle, or choose a transposition $(jn)$ and insert it into the cycle containing $j$; this gives every permutation uniquely. The full generation step distinguishes the individual class sums which this first identity groups together. Its integrality is essential: a description only over $\mathbb{Q}$ could use denominators divisible by $p$ and would not prove generation of the modular center. Since class sums form the integral basis of the center, this integral theorem reduces to the same generation statement over $F$.

Next determine their scalar action. In the characteristic-zero tableau basis of $S^{\lambda }$, $J_i$ is diagonal with entry the Content of a Young-diagram cell occupied by $i$, namely column minus row. This follows recursively along the multiplicity-free branching chain: the commuting operators preserve each path line, and $J_{i+1}=s_i{J}_i{s}_i+s_i$ determines the successive eigenvalues. Consequently a symmetric polynomial $f(J_1,\dots ,J_n)$ acts on $S^{\lambda }$ byThis is independent of the ordering of the cells. The scalar identity holds on the integral Specht lattice, so after modular reduction it depends only on the multiset of contents modulo $p$. Every composition factor has this same central character of a block.

Two simple modules are in the same block exactly when their central characters agree. The center of a block is a commutative local finite-dimensional algebra; its semisimple quotient gives one scalar character, while the orthogonal block idempotents separate different blocks. By the generation theorem, equality on all symmetric polynomials in the $J_i$ is enough. Conversely, different residue multisets are detected by elementary symmetric polynomials: the monic polynomial ${\prod }_{(i,j)\in \lambda }(z-(j-i))$ over $F$ recovers the multiplicity of each residue by unique factorization, including multiplicities larger than $p$. Thus block equivalence is exactly equality of the integersThis gives the residue-content criterion for symmetric-group blocks without assuming that the sum of contents alone distinguishes blocks.

Finally identify residue content with the core. A removable rim $p$-hook has consecutive contents modulo $p$, hence removes one cell of each residue. Therefore shapes with the same core and the same size have the same residue counts. For the converse choose $m$ divisible by $p$ and at least $n$, and write $q_r=n_r-m/p$ for the runner charges. Starting from the empty partition of an integer gives $q_r=0$. Adding a cell of residue $r$ moves a beta-number bead from residue $r-1$ to residue $r$, increasing $q_r$ and decreasing $q_{r-1}$. The same changes occur in $c_s-c_{s+1}$ when $c_r$ increases by one. Induction on the cells therefore givesEqual residue counts give equal charges and hence equal packed runner configurations, so they give the same core. This completes both directions of the block criterion. The argument exposes the three essential bridges: integral center generation, its scalar content evaluation, and the abacus identification of residue data with a unique core.

A rim hook is a connected skew diagram containing no $2\times 2$ square, whose removal leaves a partition of an integer diagram. Its leg length is its number of occupied rows minus one. The Murnaghan–Nakayama rule states that, if a permutation has cycle type $(k,\rho )$, thenIteration removes one hook for each cycle length; the empty diagram has character one at the empty cycle type.

For shape $(3,3,3)$ and cycle type $(4,3,2)$, there are exactly two removable four-cell rim hooks. The remaining shapes are $(3,2)$ and $(2,2,1)$; the removed hooks have leg lengths one and two respectively. In shape $(3,2)$ the only removable three-cell rim hook leaves $(1,1)$ and has leg length one. In shape $(2,2,1)$ the only such hook leaves $(2)$ and again has leg length one. Since the characters of $(1,1)$ and $(2)$ at a transposition are $-1$ and $1$, we obtainand henceThe cycle lengths can be taken in any order; using $4,3,2$ makes the two branches of the computation explicit.

Now consider the ordinary induction product $[\alpha ][\beta ]$. A hook partition is $(x,1^y)$, equivalently a partition of an integer whose second row has length at most one. If a hook shape $\nu$ occurs in the Littlewood–Richardson rule, then $\alpha \subseteq \nu$, forcing $\alpha$ itself to be a hook. In the skew hook $\nu /\alpha$, the new cells consist of a segment in the first row and a segment down the first column, with no shared column between those two segments. Every first-row entry must be $1$: its rightmost entry is the first letter of the lattice word, hence is $1$, and weak row increase forces all entries to its left to be $1$ too. The column segment has strictly increasing entries. Thus only $1$ can be repeated, forcing the content $\beta$ to be a hook as well. This also proves the assertion when the top segment is empty, since then every entry lies in a strictly increasing column.

Write the two nonempty hook factors as $\alpha =(a,1^k)$ and $\beta =(b,1^l)$, with $k=n-r-a$ and $l=r-b$. Let the prospective resulting hook have top-row length $x$ and leg length $y$. Its new top segment has $H=x-a$ cells and its new column segment has $V=y-k$ cells. The content consists of $b$ ones and the single letters $2,\dots ,l+1$. All those larger letters must be in the new column; its strict increase allows either no $1$ or exactly one $1$. There are therefore only two possibilities:In the first case fill the top segment with $b$ ones and the column with $2,3,\dots ,l+1$. In the second, use $b-1$ top ones and column $1,2,\dots ,l+1$. Each filling is unique and has a lattice word, including the case $b=1$ with an empty top segment in the second filling. This provesBoth displayed multiplicities are one.

To deduce the full-cycle value from this product result, rather than quote the Murnaghan–Nakayama rule again, form the virtual characterFor every proper Young subgroup $S_m\times S_{n-m}$ and every pair of its irreducible characters, the hook-product calculation givesIf a factor is not a hook there are no hook constituents; if both are hooks, their two consecutive leg lengths cancel in this alternating inner product. By Frobenius reciprocity and the completeness of the product irreducible characters, $F_n$ restricts to zero on every such subgroup. Any permutation with more than one cycle preserves a nonempty proper subset and is conjugate into one of these subgroups. Thus $F_n$ is supported only on the class of $n$-cycles.

Its inner product with the trivial character is one, since only the one-row hook contributes. That class has size $(n-1)!$, so $F_n(\rho )/n=1$, giving $F_n(\rho )=n$. For any irreducible character, the support calculation then yieldsBy character orthogonality, the left side is $(-1{)}^{n-x}$ for $\nu =(x,1^{n-x})$ and zero otherwise. These are real integers, provingFor $n=1$ the same conclusion holds directly. This derives the requested special case from the Littlewood–Richardson rule and Frobenius reciprocity.

The kernel intersection theorem for Specht modules uses maps that move entries between adjacent rows. For $1\le i<\ell (\mu )$ and $0\le v<{\mu }_{i+1}$, let $\mu (i,v)$ replace the row lengths $({\mu }_i,{\mu }_{i+1})$ by $({\mu }_i+{\mu }_{i+1}-v,v)$, keeping every other length unchanged. This is allowed to be a composition rather than a decreasing partition of an integer, and a zero row can be omitted. For a tabloid with row sets $R_j$, defineby summing over all $v$-subsets $B\subset R_{i+1}$: the new row $i+1$ is $B$, the new row $i$ is $R_i\cup (R_{i+1}\setminus B)$, and all other rows are unchanged. These are module homomorphisms, since relabelling commutes with choosing subsets. The theorem isIt holds over every field. Column cancellation shows the inclusion of $S^{\mu }$ in each kernel; the full theorem identifies the common kernel with the Specht module.

For completeness, the semistandard homomorphism theorem concerns a partition of an integer $\lambda$, a composition $\alpha$, and fillings $T$ of shape $\lambda$ with ${\alpha }_j$ occurrences of $j$. Such a filling gives ${\Theta }_T:M^{\lambda }\to M^{\alpha }$: replace its labels by the numbered entries of an input tableau, and sum the output tabloids over distinct permutations of the labels within each input row. Label $j$ assigns an entry to output row $j$. The use of distinct assignments avoids dividing by any factorial. Let ${\hat{\Theta }}_T$ be its restriction to $S^{\lambda }$. The restrictions indexed by Semistandard Young tableaux are linearly independent. They form a basis of ${\mathrm{Hom}}_{F{S}_n}(S^{\lambda },M^{\alpha })$ when $p\ne 2$, or when $\lambda$ is two-regular. In characteristic two for a two-singular shape they need not span; omitting this exception would make the statement incorrect.

We now prove the invariant vector criterion for a Specht module directly from the first theorem. The group acts transitively on the tabloid basis of $M^{\mu }$, so any invariant vector has all coefficients equal. Thus $(M^{\mu }{)}^{S_n}=F{z}_{\mu }$, where $z_{\mu }$ is the sum of all its tabloids and is nonzero over every field. Fix $i,v$, write $a={\mu }_i$, $b={\mu }_{i+1}$, and count preimages of one output tabloid. The $v$ retained entries are prescribed; among the $a+b-v$ entries of the enlarged row, exactly $a$ must have come from its old row. HenceThe common-kernel criterion becomesfor every adjacent pair, where $j=b-v$.

Suppose $p>0$, and let $t$ be least with $b<p^t$. If $a\equiv -1(\mathrm{mod}{p}^t)$, its lowest $t$ base-$p$ digits are all $p-1$. For any $0<j<p^t$, adding $j$ to $a$ creates a carry at the lowest nonzero digit of $j$. At that position the digit of $a+j$ is smaller than the corresponding digit of $j$. Lucas's theorem therefore gives $\left(\genfrac{}{}{0ex}{}{a+j}{j}\right)\equiv 0$. Conversely, if one of the first $t$ digits $a_d$ is not $p-1$, take $j=p^d$. Minimality of $t$ gives $p^d\le b$, while addition at that digit creates no carry. Lucas then yieldsThis proves the claimed equivalence, and thereforeThe invariant space, when present, is one-dimensional. At the last row put ${\mu }_{i+1}=0$; then $t_i=0$ and the condition is vacuous. This congruence formulation is for positive characteristic. In characteristic zero the positive binomial coefficients cannot vanish, so an invariant vector exists precisely for the one-row shape $(n)$.

Finally suppose $\mu$ is regular. It has a unique simple head $D^{\mu }$, by Question 1. The given sign-twisted duality of Specht modules impliesThus $D^{\mu }$ is sign exactly when the conjugate shape ${\mu }^{\prime }$ satisfies the invariant congruences.

Here is an explicit solution of those congruences under regularity. Write the positive column heights of $\mu$ as $a_1\ge \cdots \ge a_l>0$, so ${\mu }^{\prime }=(a_1,\dots ,a_l)$. Regularity says $a_i-a_{i+1}<p$, including $a_l-a_{l+1}=a_l<p$, because this difference counts rows of length $i$. Starting with $0<a_l<p$, the congruence for its preceding row forces $a_{l-1}\equiv -1(\mathrm{mod}p)$. Since $a_l\le a_{l-1}<a_l+p\le 2p-1$, the only possibility is $a_{l-1}=p-1$. Repeating upwards forces every earlier height to be $p-1$. Consequentlywhere a zero final part is omitted. Conversely this shape satisfies every invariant congruence, so its conjugate is the desired regular label of the modular sign representation:Zero parts are omitted. For $p=2$ this is $(n)$, as it should be because sign is trivial. In characteristic zero the answer is $(1^n)$.

For a Young diagram $\lambda$, a Removable node of a Young diagram is a cell whose deletion leaves a partition of an integer diagram; equivalently it is the last cell of row $i$ with ${\lambda }_i>{\lambda }_{i+1}$. Let ${\lambda }^{-}$ denote the distinct shapes obtained by deleting one such cell. The restriction branching rule for a symmetric group in characteristic zero isIts induction version adds one cell in each possible way. Over an arbitrary field, restriction instead has a Specht filtration with these successive quotients, each once; it need not split.

To prove restriction, list the removable nodes from top to bottom. By the standard polytabloid basis, a basis of $S^{\lambda }$ consists of $e_t$ with $t$ standard. Its entry $n$ is necessarily at a removable node. Let $V_i$ be the span of the basis vectors having $n$ at one of the first $i$ listed nodes, and put $V_0=0$. These are $S_{n-1}$-invariant. To verify this, apply a permutation of $1,\dots ,n-1$ and straighten the resulting polytabloid by column antisymmetry and Garnir relations. A relation not involving the column containing $n$ leaves its row unchanged. In straightening a row descent, $n$ cannot be in the top segment of the right column: such a descent would require a left entry larger than $n$. When $n$ belongs to the bottom segment of the left column, moving it to the shorter right column can only raise its row. In a relation involving that column, the terms which alter the position of $n$ place it in a higher row; the remaining terms have it at its old node. Thus straightening never moves it past the current top-to-bottom cutoff.